Answer
(a) $1\,mol\,Cr$
(b) $2.5\,mol\,O_{2}$
(c) $12\,mol\,Cr$
(d) $10.3\,mol\,O_{2}$
Work Step by Step
(a) Calculating how much product can be made based on the amount of each reactant, we have
$1\,mol\,Cr\times\frac{2\,mol\,Cr_{2}O_{3}}{4\,mol\,Cr}=0.5\,mol\,Cr_{2}O_{3}$
$1\,mol\,O_{2}\times\frac{2\,mol\,Cr_{2}O_{3}}{3\,mol\,O_{2}}=0.667\,mol\,Cr_{2}O_{3}$
Since $1\,mol\,Cr$ makes the least amount of product, it is the limiting reactant.
(b) Calculating how much product can be made based on the amount of each reactant, we have
$4\,mol\,Cr\times\frac{2\,mol\,Cr_{2}O_{3}}{4\,mol\,Cr}=2\,mol\,Cr_{2}O_{3}$
$2.5\,mol\,O_{2}\times\frac{2\,mol\,Cr_{2}O_{3}}{3\,mol\,O_{2}}=1.67\,mol\,Cr_{2}O_{3}$
Since $2.5\,mol\,O_{2}$ makes the least amount of product, it is the limiting reactant.
(c) Calculating how much product can be made based on the amount of each reactant, we have
$12\,mol\,Cr\times\frac{2\,mol\,Cr_{2}O_{3}}{4\,mol\,Cr}=6\,mol\,Cr_{2}O_{3}$
$10\,mol\,O_{2}\times\frac{2\,mol\,Cr_{2}O_{3}}{3\,mol\,O_{2}}=6.67\,mol\,Cr_{2}O_{3}$
Since $12\,mol\,Cr$ makes the least amount of product, it is the limiting reactant.
(d) Calculating how much product can be made based on the amount of each reactant, we have
$14.8\,mol\,Cr\times\frac{2\,mol\,Cr_{2}O_{3}}{4\,mol\,Cr}=7.4\,mol\,Cr_{2}O_{3}$
$10.3\,mol\,O_{2}\times\frac{2\,mol\,Cr_{2}O_{3}}{3\,mol\,O_{2}}=6.87\,mol\,Cr_{2}O_{3}$
Since $10.3\,mol\,O_{2}$ makes the least amount of product, it is the limiting reactant.